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[SOLVED] (IB18) A box contains 100 cards

karush

Well-known member
Jan 31, 2012
2,723
View attachment 1141

(a) I presume the frequency row has to equal 100
so
\(\displaystyle k=100 - (26 + 10 + 20 + 29 + 11)= 4\)


(b)(i) again presume the median is based on frequency and on ordered list
so
median of $4\ 10\ 11 \ 20\ 26\ 29 = \frac{31}{2}$ or $15.5$

(ii) interquartile range? isn't this data list 100 numbers long?
or is $Q_1=10$ and $Q_3=26$ so interquartile range$=26-10=13$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: (IB18) A box contains a 100 cards

For part b), you want to use the data, not the frequencies, in your calculations. To find the median (or $Q_2$) you observe that there is an even number of elements, so you take the arithmetic mean of the 50th and 51st elements.

Now, since there is an even number of elements in each half, you want to take the arithmetic mean of the 25th and 26th elements as $Q_1$ and the arithmetic mean of the 75th and 76th elements as $Q_3$. And then the inter-quartile range is given by:

\(\displaystyle IQR=Q_3-Q_1\)
 

karush

Well-known member
Jan 31, 2012
2,723
Re: (IB18) A box contains a 100 cards

For part b), you want to use the data, not the frequencies, in your calculations. To find the median (or $Q_2$) you observe that there is an even number of elements, so you take the arithmetic mean of the 50th and 51st elements.

Now, since there is an even number of elements in each half, you want to take the arithmetic mean of the 25th and 26th elements as $Q_1$ and the arithmetic mean of the 75th and 76th elements as $Q_3$. And then the inter-quartile range is given by:

\(\displaystyle IQR=Q_3-Q_1\)
I got 5-1=4 IQR
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775